Question

# A stone is thrown vertically upward with an initial velocity of $40m{s}^{-1}$. Taking $g=10m{s}^{-2}$, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Open in App
Solution

## Step 1: GivenInitial velocity (u) = $40m{s}^{-1}$ (in upward direction)Final velocity (v) = $0m{s}^{-1}$Acceleration due to gravity, $g=10m{s}^{-2}$Step 2: Formula used and calculation${v}^{2}={u}^{2}+2as\phantom{\rule{0ex}{0ex}}0={40}^{2}+2×\left(-10\right)×s$(-ve sign is used because the stone is going upward)$20s=1600\phantom{\rule{0ex}{0ex}}s=\frac{1600}{20}\phantom{\rule{0ex}{0ex}}s=80m$Hence, the maximum height is $80m$.Step 3: Calculation of net displacement and total distance coveredTotal distance covered = $2×$maximum height $=2×80m\phantom{\rule{0ex}{0ex}}=160m$Net displacement is zero because initial position and final position are same.

Suggest Corrections
56
Explore more